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\textsc{中国科学院大学, 计算机与控制学院} \\ [25pt] % Your university, school and/or department name(s)
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\huge 随机过程第二次作业 \\ % The assignment title
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\author{黎吉国} % Your name

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\begin{document}

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%	PROBLEM 1
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\section{随机过程相关函数的计算}
\textbf{1.} 设随机过程$\{X(t),t\inT\}$由下述三个样本函数组成且等可能发生：
\[ X(t,\omega_1) = 1,\quad  X(t,\omega_2) = A\sin{t},\quad  X(t,\omega_3) = \cos {t}\]
其中$A$是常数，试计算此过程的：(1)均值函数$m_X(t)$;\ (2)相关函数$R_X(s,t)$.\\
\textbf{解：}\\
(1)均值函数是在$\omega$维度上求$X(t)$的均值：$m_X(t)=E(X(t))=\frac{1}{3}(1+A\sin{t}+\cos{t})$.\\
(2)$R_X(s,t)=E(X(s)X(t))$，该随机过程$\{X(t),t\in T\}$只有三个样本函数，故：
\[ P\{X(s)=1, X(t)=1\}=\frac{1}{3},\quad P\{X(s)=A\sin {s},X(t)=\sin{t}\}=\frac{1}{3}, \quad P\{X(s)=\cos{s},X(t)=cos{t}\}=\frac{1}{3}\]
其余情况概率都为$0$.所以有:
\[ R_X(s,t)=E(X(s)X(t))=\frac{1}{3}(1+A^2 \sin{s}\sin{t}+\cos{s}\cos{t}) \]
{\color{red}错误解答：}\\
$X(s)X(t)$的复合有9中情况，分别是
\begin{align*}
X(s,\omega_1)X(t,\omega_1),X(s,\omega_1)X(t,\omega_2),X(s,\omega_1)X(t,\omega_3) \\
X(s,\omega_2)X(t,\omega_1),X(s,\omega_2)X(t,\omega_2),X(s,\omega_2)X(t,\omega_3)\\
X(s,\omega_3)X(t,\omega_1),X(s,\omega_3)X(t,\omega_2),X(s,\omega_3)X(t,\omega_3)
\end{align*}\\
所以该随机过程的相关是：
\[R_X(s,t)=\frac{1}{9}(1+A\sin{s}+\cos{s}+A\sin{t}+A^2\sin{t}\sin{s}+A\sin{t}\cos{s}+\cos{t}+A\cos{t}\sin{s}+\cos{t}\cos{s}\]
{\color{red}错误原因：}没有理解随机过程的相关是在$\omega$维度上的均值，研究的是同一事件在不同时间的关系。

\section{均方极限的概念}
\textbf{2.}设$\{X_n,n\ge 1\}$是相互独立的随机变量序列，其分布律为
\begin{equation*}
X\sim\left(
\begin{array}{cc}
n&0\\
\frac{1}{n^2}&1-\frac{1}{n^2}
\end{array}
\right)
\end{equation*}\\
试讨论此序列是否均方收敛。\\
\textbf{解：}\\
Cauchy准则：若$\{X_n,n=1,2,\ldots\}\subset H$，则$\{X_n,n=1,2,\ldots\}$均方收敛的充要条件是
\[\lim_{m,n\to +\infty}E(|X_m-X_n|^2)=0\]
只需求出$E(|X_m-X_n|^2)=E(X_m^2)+E(X_n^2)-2E(X_m)E(X_n)$的极限。\\
其中
\[E(X_m)=m\cdot \frac{1}{m^2}=\frac{1}{m},\quad E(X_m^2)=m^2\cdot\frac{1}{m^2}=1\]
\[E(X_n)=n\cdot \frac{1}{n^2}=\frac{1}{n},\quad E(X_n^2)=n^2\cdot \frac{1}{n^2}=1\]
当$m\ne n$时，有
\[E(X_mX_n)=E(X_m)E(X_n)=\frac{1}{mn}\]
则
\[\lim_{m,n\to +\infty}E(|X_m-X_n|^2)=\lim_{m,n\to +\infty}(2-2\frac{1}{mn})=2\]
所以，该随机序列不收敛。\\
{\color{red}入选理由：}该题向我们展示了如何借助极限的性质来证明一个序列的收敛性。
\section{宽平稳的概念}
\textbf{3.}设有随机过程$\{X(t)=\sin{\omega t},-\infty<t<\+\infty\}$，其中$\omega \sim U(0,2\pi)$，试证$\{X(n),n=0,1,2,\ldots\}$是宽平稳随机序列。\\
\textbf{证明}\\
\textbf{宽平稳：}对于随机过程$X(t),t\in T$，如果$\any t,s\in T$，都有
\[E(X(t))=E(X(s))\]
\[R_X(t,s)=R_X(t+D,s+D)\qquad \any D\in T\]
则称随机过程$X(t)$具有宽平稳性。用文字语言描述就是随机过程的均值是常数，其自相关函数只依赖于时间差$t-s$。\\
只需证明其均值是常数，自相关函数只依赖于相对时间即可。\\
\[m_X(n)=E(X(n))=\int_0^{2\pi}\frac{1}{2\pi}\sin{n\omega}d\omega = 0\]
\begin{align*}
R_X(n,n+k) &= E(X(n)X(n+k))\\
&=\int_0^{2\pi}\frac{1}{2\pi}\sin{n\omega}\sin(n+k)\omega d\omega\\
&=\frac{1}{2\pi}\cdot \frac{1}{2}\int_0^{2\pi}[\cos k\omega-\cos(2n+k)\omega]d\omega\\
&=
\begin{cases}
  \frac{1}{2},\quad k=0\\
  0,\quad k\ne 0
\end{cases}
\end{align*}
故该序列是宽平稳随机序列。\\
{\color{red}入选理由：}该题向我们展示了如何证明一个随机序列的宽平稳性。

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